Find the sum to n terms of the series 3 times | vedclass

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(C) The $n^{	ext{th}}$ term $a_n$ of the series is given by the product of the $n^{	ext{th}}$ terms of the two sequences $(3, 6, 9, \dots)$ and $(8,

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$3n(n+1)(n+3)$

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(C) The $n^{	ext{th}}$ term $a_n$ of the series is given by the product of the $n^{	ext{th}}$ terms of the two sequences $(3, 6, 9, \dots)$ and $(8, 11, 14, \dots)$.$a_n = (3n) 	imes (3n + 5) = 9n^2 + 15n$.To find the sum $S_n = \sum_{k=1}^n a_k$,we calculate:$S_n = \sum_{k=1}^n (9k^2 + 15k) = 9 \sum_{k=1}^n k^2 + 15 \sum_{k=1}^n k$.Using the standard summation formulas $\sum k^2 = \frac{n(n+1)(2n+1)}{6}$ and $\sum k = \frac{n(n+1)}{2}$:$S_n = 9 	imes \frac{n(n+1)(2n+1)}{6} + 15 	imes \frac{n(n+1)}{2}$.$S_n = \frac{3n(n+1)(2n+1)}{2} + \frac{15n(n+1)}{2}$.$S_n = \frac{3n(n+1)}{2} (2n + 1 + 5) = \frac{3n(n+1)}{2} (2n + 6)$.$S_n = 3n(n+1)(n+3)$.

Find the sum to $n$ terms of the series $3 \times 8 + 6 \times 11 + 9 \times 14 + \dots$

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